# PharmPK Discussion - Polynomial regression

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• On 9 Mar 2004 at 14:07:14, "Daniel Rossi de Campos" (Daniel.Campos.at.anvisa.gov.br) sent the message
`Dear All,Sorry, but a need to make this basic query:Could anyone say how to calculate the concentrations from a polinomialregression (Y = ax2 + bx + c where: y = area ratio and x =concentration (ng/mL) in a bioanalytical assay?Thanks,DanielP.S n I have tried to use Baskara theorema, but I could not getadequate results.`
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• On 9 Mar 2004 at 16:52:11, Richard Sams (sams.at.mail.atl.ohio-state.edu) sent the message
`Daniel,You can perform this calculation easily in an Excel spreadsheet afteryou have obtained the polynomial expression that best describes therelationship between the area ratio and the concentrations of yourcalibrators by using the "Goal Seek" function in the drop down menuunder "Tools". You will be asked to input the following:"Set cell:" You need to enter the designation of the cell in which thepolynomial equation is expressed"To value:" You need to enter the numerical value of the area ratio"By changing cell:" You need to enter the designation of the cell inwhich you want the concentration to be reported.After you have provided this information, click on "OK" in the dialogbox and the concentration will appear in the cell that you havedesignated for the answer.Sincerely,Richard Sams`
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• On 10 Mar 2004 at 08:45:59, Michael.Gassen.at.4sc.com sent the message
`Dear Daniel,For a square polynomal function, you have a simple option in Excel:1. Make a graph of your data (x-y type)2. Add a trendline, use the option "put equation in graph"For anything more complicated, you can use the solver add-in (needsinstallation via add-in manager), which will do numerical regression forbasicly every equation. You just make a spreadsheet with your measureddataY. For e.g. an equation like  Y = ax3 + bx2 + cx + d, you create foucellswhere you type your guestimates for the parameters a-d. Based on theseparameters and your equation, you calculate the "theoretical" values ofY(th) and the sum of square differences sum((Y-Y(th))^2). Then youminimizethis sum with the solver with variation of the parameters (a-d).good luck,MichaelMichael GassenHead of Pharmacology and Preclinical Development4SC AG                           Fon: +49 89 700763-0Am Klopferspitz 19a         Fax: +49 89 700763-2982152 Martinsried            E-Mail: gassen.-a-.4sc.comGermany                          Internet: www.4sc.com`
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• On 10 Mar 2004 at 10:50:37, Pascal.Delrat.-a-.uk.netgrs.com sent the message
`Hi,since C and Y are constant in your eqution,why not try to resolve ax2 + bx + c -Y =0You should end up on ax2 + bx + K =0 with K =c-Y since Y should be > tocyou should end up on :ax2 + bx - K =0Then resolve the equation for x with finding the two roots with onemight benegative (that you reject) and the other one your concentration.It would help to have the constants of your equation to know whichsolutionis adequate.Do not if this helps,Regards,Pascal`
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• On 10 Mar 2004 at 10:55:10, David_Critchley.aaa.eisai.net sent the message
`Dear Daniel,There will be 2 outcomes possible for such a quadratic function (sousing'goal seek' may not be the best way to solve) - obviously, only one willcorrespond to your concentration (you will know which is correct).  Toworkout the two possible values for x, use the following formulae (where Y= ax2 + bx + c):x1 = - (b/(2a)) - {square root [ ((Y-c)/a) + (b/(2a))2]}x2 = - (b/(2a)) + {square root [ ((Y-c)/a) + (b/(2a))2]}I hope this is helpful.Cheers,Dave`
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• On 15 Mar 2004 at 10:17:05, "Henri Merdjan" (hmerdjan.-a-.wanadoo.fr) sent the message
`Dear members,I am writing regarding a recent debateabout the use of polynomials forcalibration purpose. Let me add a few comments on it.I haven't retained the original messages. However, if my recollectionis correct, several members provided the right answers. Morespecifically, two solutions exist for ax2 + bx +c =0. By the way,somebody wrote that one is positive and the other is negative. This isnot always right, both may be positive, or both negative.The existence of 2 solutions illustrate an undesirable property ofpolynomials in calibrations: they are non-monotonic functions. For thisreason, I would like to suggest 2 alternatives. Bothare basedonmonotonic functions.1. One optionwould beto use linear regression on log-transformationdata (on both axes). LnY=a*LnX + b. This option, however,suffers 2main (hidden) drawbacks. One is an underlying assumption of aconstant-CV error model, rarely applicable to bioanalytical data.Another one is that the model is essentially forced through the origin.This is obvious after antilog transformation: Y = exp(b)*X**a.Themain value of this approachis its straightforward implementation.2. The second option would be tu use NONlinear regression onUNtransformed data,usingthe monotonic model: Y = a*X**n + b. Bycontrast with option 1, these is no specific assumption about the errormodel (feel free to use the most appropriate weighting function, ifany) and the model is not forced through the origin thanks to theintercept "b". On the other hand, implementation is not straightforwardasit requires a nonlinear regression module (e.g. Excel solver).Enjoy the testing!HenriHenri MerdjanConsultant to the Life Science IndustriesPK and PK/PD specialist34, rue des Bergers75015 ParisFrancehmerdjan.-at-.wanadoo.fr`
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• On 18 Mar 2004 at 09:55:43, "J.H.Proost" (J.H.Proost.at.farm.rug.nl) sent the message
`Dear Henri,Thank your for your suggestions for alternatives for the polynomialregression.Your comments with respect to the properties of linear regressionapplied to log-transformed data are correct, but I do not fully agreewith your conclusions:- the underlying assumption of a constant-CV error may be not fullyjustified, but this also applies to the assumption of a constant errorunderlying the use of linear regression on untransformed and unweighteddata. In real data, the situation is usually somewhere between.- forcing the model through the origin is not a bad idea. In case oneknows that there a blank sample produces a response, one can makecorrections for it. Anyhow, if this is the case, also linear regressionmay give rise to inaccuracies, and one has to be quite sure that thecalibration curve is linear. Please note that after log-transformationthe concentration zero is infinitely far away from the lowestcalibration sample. Nobody would ever make such an extrapolation! So itis not really important that the calibration curve in the untransformedscale is forced through the origin.In addition, linear regression requires that the X-values are regularlydistributed over the range. This is usually not the case. Afterlogarithmic transformation this requirement may be fulfilled better (inpractice, again, the situation is probably somewhere between).With respect to the use NONlinear regression and the monotonic model: Ya*X**n + b: If data are not weighted, there is certainly a specifiedunderlying (hidden) assumption about the error model. I would notrecommend to use this method without careful analysis of the results,e.g. in residuals plots, quite similar to common practice inpharmacokinetics.Best regards,HansJohannes H. ProostDept. of Pharmacokinetics and Drug DeliveryUniversity Centre for PharmacyAntonius Deusinglaan 19713 AV Groningen, The Netherlandstel. 31-50 363 3292fax  31-50 363 3247Email: j.h.proost.-a-.farm.rug.nl`
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